SSC JE Electrical Engineering 2014 Set 2 Paper

Question : 196 of 200

Marks: +1, -0

Thevenin's equivalent voltage and resistance between the terminal A and B for network of given figure is :

Solution:

R_{\TH}=5∥ 15={5×15}/{5+15}

=75/20=3.75 Ω

For thevenin voltage

V_{\TH}

,
According to Kirchoff's voltage law:
From Loop 1

5 I_1 + V_{\TH} -10 = 0

...(i)
From Loop 2

15 I1 + 20 -V_{\TH} = 0

...(ii)
Solving equations (i) and (ii) we get

15({10-V_{\TH}/5)+20-V_{\TH}=0

∴

30 -3V _{\TH} + 20 -V_{\TH} = 0

∴

V_{\TH}=50/4= 12.5 V

∴ Option (c) is correct.

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